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Database Systems

October 15, 2008Lecture #5

Course Administration

• Assignment #2 is out on the course homepage.– It is due in two weeks 10.29.2008.

• Assignment #1 solution will be on the course homepage.• Next week reading:

– Chapter 8: Overview of Storage and Indexing

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Bump Top (U. of Toronto)

• A virtual desktop is like a real desktop

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Long Reflection: DB design

• Step 1: Requirements Analysis– What data to store in the database?

• Step 2: Conceptual Database Design– Come up with the design: Entity-Relation (ER) model – Sketch the design with ER diagrams

• Step 3: Logical Database Design– Implement the design: relational data model– Map ER diagrams to relational tables

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Recent Reflection: DB design

• Last lecture:– Query language: how to ask questions about the [relational] database?– Mathematical query language: Relational Algebra

• This lecture – A real query language: SQL (Structured Query Language)

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Review: Relational Algebra

• A query is applied to table(s), and the result of a query is also a table.

• Find the names of sailors who have reserved boat 103πsname((σbid = 103 Reserves) ∞ Sailors)

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Example Table Definitions

Sailors(sid: integer, sname: string, rating: integer, age: real)Boats(bid: integer, bname: string, color: string)Reserves(sid: integer, bid: integer, day: date)

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Review: Relational Algebra• Basic relational algebra operators:

– Selection (σ, pronounced sigma): Select a subset of rows from a table.

– Projection (π): Delete unwanted columns from a table.– Cross-product ( X ): Combine two tables.– Set-difference ( - ): Tuples in table 1, but not in table 2.– Union ( U ): Tuples in tables 1 or 2.

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Review: Relational Algebra (more)

• Additional relational algebra operators:– Intersection (∩) : tuples in both tables 1 and 2.– Join (∞): conditional cross product– Division (/)– Renaming (p)

• Operations composed into complex query expression• Query in English?

πsid (σ age > 20 Sailors) –

πsid ((σ color = ‘red’ Boats) ∞ Reserves ∞ Sailors)

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Relational Algebra to SQL

• Relational operators → SQL commandsRelational Algebra:

πsname (σbid = 103 (Sailors∞ Reserves))

SQL:SELECT S.sname

FROM Sailors S, Reserves RWHERE S.sid=R.sid AND R.bid=103

• Guess the mapping?– Notice the difference between SELECT (SQL) and σ

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SQL: Queries, Constraints, Triggers

Chapter 5

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Lecture Outline

• Basic Query– SELECT

• Set Constructs– UNION, INTERSECT,

EXCEPT, IN, ANY, ALL, EXISTS

• Nested Queries• Aggregate Operators

– COUNT, SUM, AVG, MAX, MIN, GROUP BY, HAVING

• Null Values• Integrity Constraints

– CHECK, CREATE ASSERTION• Triggers

– CREATE TRIGGER, FOR EACH ROW

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Example Table Definitions

Sailors(sid: integer, sname: string, rating: integer, age: real)Boats(bid: integer, bname: string, color: string)

Reserves(sid: integer, bid: integer, day: date)

• Find names of sailors who’ve reserved boat #103SELECT S.sname

FROM Sailors S, Reserves RWHERE S.sid=R.sid AND R.bid=103

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Basic SQL Query

SELECT [DISTINCT] target-listFROM relation-listWHERE qualification

• Relation-list: A list of relation names (possibly with range-variable after each name).

• Target-list: A list of attributes of relations in relation-list• Qualification: conditions on attributes (<, >, =, and, or,

not, etc.)• DISTINCT: optional keyword for duplicate removal.

– Default = no duplicate removal!

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How to evaluate a query?

SELECT [DISTINCT] target-listFROM relation-listWHERE qualification• Conceptual query evaluation using relational operators:

1) Compute the cross-product of relation-list.2) Discard resulting tuples if they fail qualifications.3) Delete attributes that are not in target-list. (called column-list)4) If DISTINCT is specified, eliminate duplicate rows.

• Only conceptual because of inefficiency computation– An optimizer can find better strategy

SELECT S.snameFROM Sailors S, Reserves R

WHERE S.sid=R.sid AND R.bid=103

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Example of Conceptual Evaluation (1)

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.5

58 rusty 10 35.0

(1) Compute the cross-product of relation-list.

SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid=R.sid AND R.bid=103

Sailors Reserves

sid bid day22 101 10/10/9658 103 11/12/96

X

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Example of Conceptual Evaluation (2)

(2) Discard tuples if they fail qualifications.

S.sid sname rating age R.sid bid day

22 dustin 7 45.0 22 101 10/10/96

22 dustin 7 45.0 58 103 11/12/96

31 lubber 8 55.5 22 101 10/10/96

31 lubber 8 55.5 58 103 11/12/96

58 rusty 10 35.0 22 101 10/10/96

58 rusty 10 35.0 58 103 11/12/96

SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid=R.sid AND R.bid=103

Sailors X Reserves

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Example of Conceptual Evaluation (3)

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 22 101 10/10/96

22 dustin 7 45.0 58 103 11/12/96

31 lubber 8 55.5 22 101 10/10/96

31 lubber 8 55.5 58 103 11/12/96

58 rusty 10 35.0 22 101 10/10/96

58 rusty 10 35.0 58 103 11/12/96

(3) Delete attribute columns that not in target-list.

Sailors X Reserves

SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid=R.sid AND R.bid=103 sname

rusty

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A Note on Range Variables

• Really needed range variables only if the same relation appears twice in the FROM clause.

SELECT S.snameFROM Sailors as S, Reserves RWHERE S.sid=R.sid AND bid=103

SELECT snameFROM Sailors, Reserves WHERE Sailors.sid=Reserves.sid AND bid=103

OR

SELECT snameFROM Sailors S, Reserves R1, Reserves R2 WHERE S.sid = R1.sid AND S.sid = R2.sid AND

R1.bid <> R2.bid

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Find the sids of sailors who’ve reserved at least one boat

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 22 101 10/10/96

22 dustin 7 45.0 58 103 11/12/96

31 lubber 8 55.5 22 101 10/10/96

31 lubber 8 55.5 58 103 11/12/96

58 rusty 10 35.0 22 101 10/10/96

58 rusty 10 35.0 58 103 11/12/96

Sailors X Reserves

SELECT S.sidFROM Sailors S, Reserves RWHERE S.sid=R.sid

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DISTINCT

• Find the names and ages of all sailorsSELECT S.sname, S.ageFROM Sailors S

• Add DISTINCT to this query?• Replace S.sname by S.sid in the

SELECT clause? • Add DISTINCT to the above?

Sid Sname Rating Age

22 Dustin 7 45.0

29 Brutus 1 33.0

31 Lubber 8 55.5

32 Andy 8 25.5

58 Rusty 10 35.0

64 Horatio 7 35.0

71 Zorba 10 16.0

74 Horatio 9 35.0

85 Art 3 25.5

95 Bob 3 63.5

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Find sailors whose names begin and end with B and contain at least three characters.

SELECT S.age, age1=S.age-5, 2*S.age AS age2

FROM Sailors SWHERE S.sname LIKE ‘B_%B’• AS and = are two ways to

name fields in result.• LIKE for string matching.

– `_’ for one character – `%’ for 0 or more characters.

Sid Sname Rating Age22 Dustin 7 45.029 Brutus 1 33.031 Lubber 8 55.574 Horatio 9 35.085 Art 3 25.595 Bob 3 20

Age Age1 Age220 15 40

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Find sid’s of sailors who’ve reserved a red or a green boats.

SELECT DISTINCT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND (B.color=‘red’ OR B.color=‘green’)

• UNION: work on two union-compatible sets of tuples SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’UNIONSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

Find sid’s of sailors who’ve reserved a red and a green boat

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’INTERSECTSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

• (A Except B) returns tuples in A but not in B.• What is the query in English if we replace INTERSECT by EXCEPT?

– Find sids of all sailors who have reserved a red boat but not a green boat.

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SET Construct: UNION ALL

• UNION, INTERSECT, and EXCEPT delete duplicate by default.

• To retain duplicates, use UNION ALL, INTERSECT ALL, or EXCEPT ALL.

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Sid Sname71 Zorba74 Horatio74 Horatio95 Bob

Sid Sname22 Dustin71 Zorba74 Horatio74 Horatio

INTERSECT ALL

=

Sid Sname71 Zorba74 Horatio74 Horatio

Nested Queries

• WHERE clause can contain an SQL subquery.– (Actually, so can FROM and HAVING clauses.)

• Find names of sailors who’ve reserved boat #103:SELECT S.snameFROM Sailors SWHERE S.sid IN (SELECT R.sid FROM Reserves R WHERE R.bid=103)

• (x IN B) returns true when x is in set B.– To find sailors who’ve not reserved #103, use NOT IN.

• Nested loops Evaluation– For each Sailors tuple, check the qualification by computing the subquery.– Does the subquery result change for each Sailor row?

• When would subquery result change for each Sailor row?

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Subquery: finds sidswho have reservedbid 103

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Nested Queries with CorrelationSELECT S.snameFROM Sailors SWHERE EXISTS (SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid )

• EXISTS is another set operator, like IN. – (EXISTS S) returns true when S is not empty.

• What is the above query in English?– Find sailors who have reserved boat #103

• In case of correlation, subquery must be re-computed for each Sailors tuple.

Correlation: subquery finds all reservations forbid 103 from current sid

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Nested Queries with UNIQUE

Sailors(sid: integer, sname: string, rating: integer, age: real)Boats(bid: integer, bname: string, color: string)

Reserves(sid: integer, bid: integer, day: date)

• (UNIQUE S) returns true if S has no duplicate tuples or S is empty.SELECT S.snameFROM Sailors SWHERE UNIQUE (SELECT R.bid FROM Reserves R WHERE R.bid=103 AND S.sid=R.sid)

• What is the above query in English?– Finds sailors with at most one reservation for boat #103.

• Replace R.bid with *?– Finds sailors with at most one reservation for boat #103 in a given day.

Reservessid bid day

22 101 10/10/96

58 103 11/12/96

58 103 12/12/96

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More on Set-Comparison Operators• Have seen IN, EXISTS and UNIQUE. Can also use NOT IN, NOT EXISTS, and

NOT UNIQUE.• Also available: op ANY, op ALL, where op can be >, <, =, ≠, ≤, ≥

– (a > ANY B) returns true when a is greater than any one element in set B.– (a > ALL B) returns true when a is greater than all elements in set B.SELECT *FROM Sailors SWHERE S.rating > ANY (SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)

• What is the above query in English?– Find sailors whose rating is greater than that of some sailor called Horatio.

• What is the above query in English if > ANY is replaced by > ALL?– Find sailors whose rating is greater than all sailors called Horatio.

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Find sid’s of sailors who’ve reserved a red and a green boat

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’INTERSECTSELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’

• Rewrite INTERSECT with IN (plus a subquery)– (x IN B) returns true when x is in set B.– Strategy?

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Rewriting INTERSECT Using IN

• Find sid’s of Sailors who’ve reserved red but not green boats (EXCEPT)– Replace IN with NOT IN.

SELECT S.sidFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ AND S.sid IN (SELECT S2.sid FROM Sailors S2, Boats B2, Reserves R2 WHERE S2.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)Find sids who’ve reserved a green boat

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Division in SQL• Find sailors who’ve

reserved all boats.• Strategy?

– Find all boats that have been reserved by a sailor

– Compare with all boats– Do the sailor’s reserved

boats include all boats?• Yes → include this sailor• No → exclude this sailor

SELECT S.snameFROM Sailors SWHERE NOT EXISTS ((SELECT B.bid FROM Boats B) EXCEPT (SELECT R.bid FROM Reserves R WHERE R.sid=S.sid))

(A EXCEPT B) returns tuples in A but not in B.

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Division in SQL

• Can you do it the hard way, without EXCEPT & with NOT EXISTS?

• Strategy:– For each sailor, check that there is no boat

that has not been reserved by this sailor.SELECT S.snameFROM Sailors SWHERE NOT EXISTS ( SELECT B.bid FROM Boats B WHERE NOT EXISTS ( SELECT R.bid

FROM Reserves R WHERE R.bid = B.bid AND R.sid = S.sid))

bid bname color

101 xyz red

103 abc green

Boats

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.5

Sailors

Reserves

sid bid day

22 101 10/10/96

31 101 11/12/96

31 103 12/12/96

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Aggregate Operators

• COUNT (*)• COUNT ( [DISTINCT] A)

– A is a column• SUM ( [DISTINCT] A)• AVG ( [DISTINCT] A)• MAX (A)• MIN (A)• Count the number of sailors

SELECT COUNT (*)FROM Sailors S

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Find the average age of sailors with rating = 10

Sailors(sid: integer, sname: string, rating: integer, age: real)

SELECT AVG (S.age)FROM Sailors SWHERE S.rating=10

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Count the number of different sailor names

Sailors(sid: integer, sname: string, rating: integer, age: real)

SELECT COUNT (DISTINCT S.sname)FROM Sailors S

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Find the age of the oldest sailor

Sailors(sid: integer, sname: string, rating: integer, age: real)

SELECT MAX(S.AGE)FROM Sailors S

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Find name and age of the oldest sailor(s)

SELECT S.sname, MAX (S.age)FROM Sailors S

• This is illegal, but why?– Cannot combine a column with a value (unless we use

GROUP BY)SELECT S.sname, S.age

FROM Sailors S WHERE S.age = (SELECT MAX (S2.age) FROM Sailors S2)

• Okay, but not supported in every system– Convert a table (of a single aggregate value) into a

single value for comparison

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GROUP BY and HAVING

• So far, aggregate operators are applied to all (qualifying) tuples. – Can we apply them to each of several groups of tuples?

• Example: find the age of the youngest sailor for each rating level.

– In general, we don’t know how many rating levels exist, and what the rating values for these levels are!

– Suppose we know that rating values go from 1 to 10; we can write 10 queries that look like this:

SELECT MIN (S.age)FROM Sailors SWHERE S.rating = i

For i = 1, 2, ... , 10:

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Find the age of the youngest sailor for each rating level

SELECT S.rating, MIN (S.age) as ageFROM Sailors SGROUP BY S.rating

(1) The sailors tuples are put into “same rating” groups.

(2) Compute the Minimum age for each rating group.

Sid Sname Rating Age22 Dustin 7 45.031 Lubber 8 55.585 Art 3 25.532 Andy 8 25.595 Bob 3 63.5

Rating Age3 25.53 63.57 45.08 55.58 25.5

Rating Age3 25.57 45.08 25.5

(1)

(2)

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Find the age of the youngest sailor for each rating level that has at least 2 members

SELECT S.rating, MIN (S.age) as minage

FROM Sailors SGROUP BY S.ratingHAVING COUNT(*) > 1

1. The sailors tuples are put into “same rating” groups.

2. Eliminate groups that have < 2 members.

3. Compute the Minimum age for each rating group.

Sid Sname Rating Age22 Dustin 7 45.031 Lubber 8 55.585 Art 3 25.532 Andy 8 25.595 Bob 3 63.5

Rating Age3 25.53 63.57 45.08 55.58 25.5

Rating Minage3 25.58 25.5

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Queries With GROUP BY and HAVING

SELECT [DISTINCT] target-listFROM relation-listWHERE qualificationGROUP BY grouping-listHAVING group-qualification

• The target-list contains (i) attribute names (ii) terms with aggregate operations (e.g., AVG (S.age)).

• The attribute list (e.g., S.rating) in target-list must be in grouping-list.

• The attributes in group-qualification must be in grouping-list.

SELECT S.rating, MIN (S.age) as ageFROM Sailors SGROUP BY S.ratingHAVING S.rating > 5

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Say if Attribute list is not in grouping-list

SELECT S.sname, S.rating, AVG (S.age) as age

FROM Sailors SGROUP BY S.rating HAVING COUNT(S.rating) > 1

Sname Rating AgeArt 3 25.5Bob 3 63.5Dustin 7 45.0Lubber 8 55.5Andy 8 25.5

Sname Rating Age? 3 44.5? 8 40.5

Sid Sname Rating Age22 Dustin 7 45.031 Lubber 8 55.585 Art 3 25.532 Andy 8 25.595 Bob 3 63.5

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Say if Group qualification is not in grouping-list

SELECT S.rating, AVG (S.age) as age

FROM Sailors SGROUP BY S.rating HAVING S.sname ≠ ‘Dustin’

Sname Rating AgeArt 3 25.5Bob 3 63.5Dustin 7 45.0Lubber 8 55.5Andy 8 25.5

Rating Age ?

Sid Sname Rating Age22 Dustin 7 45.031 Lubber 8 55.585 Art 3 25.532 Andy 8 25.595 Bob 3 63.5

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Conceptual Evaluation

• Without GROUP BY and HAVING: – Compute cross-product of relation-list– Remove tuples that fail qualification– Delete unnecessary columns

• With GROUP BY and HAVING, continue with– Partition remaining tuples into groups by the value of attributes in

grouping-list (specified in GROUP-BY clause)– Remove groups that fail group-qualification (specified in HAVING

clause). – Compute one answer tuple per qualifying group.

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For each red boat, find the number of reservations for this boat

SELECT B.bid, COUNT (*) AS num_reservations

FROM Boats B, Reserves RWHERE R.bid=B.bid AND

B.color=‘red’GROUP BY B.bid

SELECT B.bid, COUNT (*) AS num_reservations

FROM Boats B, Reserves RWHERE R.bid=B.bid GROUP BY B.bidHAVING B.color=‘red’

• Illegal, why?– B.color does not appear in

group-list

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Find the age of the youngest sailor with age > 18 for each rating with at least 2 sailors (of any age)

SELECT S.rating, MIN (S.age)

FROM Sailors SWHERE S.age > 18GROUP BY S.ratingHAVING COUNT(S) > 1

• What is wrong?– COUNT(S) is counting

tuples after the qualification (S.age > 18).

– Eliminate groups with multiple sailors but only one sailor with age > 18.

• How to fix it?– Use subquery in the HAVING

clause.

SELECT S.rating, MIN (S.age)FROM Sailors SWHERE S.age > 18GROUP BY S.ratingHAVING 1 < ANY (SELECT COUNT (*) FROM Sailors S2 WHERE

S.rating=S2.rating)

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Find rating(s) for (which the average age is the minimum) over all rating groups

SELECT S.ratingFROM Sailors SWHERE S.age = (SELECT MIN (AVG (S2.age)) FROM Sailors S2 GROUP BY S2.rating)

• What’s wrong? – Aggregate operations

cannot be nested

• How to fix it?

SELECT Temp.ratingFROM (SELECT S.rating, AVG

(S.age) AS avgage FROM Sailors S GROUP BY S.rating) AS

TempWHERE Temp.avgage = (SELECT

MIN (Temp.avgage) FROM Temp)

A temp table (rating, avg age)

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Table Constraints

• Specify constraints over a single table– Useful when more general

ICs than keys are involved.

• Constraints can be named.

CREATE TABLE Sailors( sid INTEGER,sname CHAR(10),rating INTEGER,age REAL,PRIMARY KEY (sid),CHECK ( rating >= 1

AND rating <= 10 ) CREATE TABLE Reserves

( sname CHAR(10),bid INTEGER,day DATE,PRIMARY KEY (bid,day),CONSTRAINT noInterlakeResCHECK (`Interlake’ ≠

( SELECT R.bnameFROM Reservers RWHERE R.bid=bid)))

The boat ‘Interlake’ cannot be reserved

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Assertions: Constraints Over Multiple Tables

CREATE TABLE Sailors( sid INTEGER,sname CHAR(10),rating INTEGER,age REAL,PRIMARY KEY (sid),CHECK ( (SELECT COUNT (S.sid) FROM Sailors S)+ (SELECT COUNT (B.bid) FROM Boats B) < 100 )

• Awkward and wrong!– If Sailors is empty,

the number of Boats tuples can be anything!

• ASSERTION is the right solution; not associated with either table.

CREATE ASSERTION smallClubCHECK ( (SELECT COUNT (S.sid) FROM Sailors S)+ (SELECT COUNT (B.bid) FROM Boats B) < 100 )

Number of boatsplus number of sailors is < 100

Triggers

• Trigger: procedure that starts automatically if specified changes occur to the DBMS

• A trigger has three parts:– Event (activates the trigger)– Condition (tests whether the triggers should run)– Action (what happens if the trigger runs)

CREATE TRIGGER incr_count AFTER INSERT ON Students // EventWHEN (new.age < 18) // Condition FOR EACH ROW

BEGIN // ACTION: a procedure in Oracle’s PL/SQL syntaxcount := count + 1

END

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Starwar Exercises

char(name, race, homeworld, affiliation)planets(name, type, affiliation)

timetable(cname, pname, movie, arrival, departure)

Which planet does Princess Leia go to in movie3?

SELECT distinct pname FROM timetableWHERE cname ='Princess Leia' and movie=3

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Starwar Exercises

char(name, race, homeworld, affiliation)planets(name, type, affiliation)

timetable(cname, pname, movie, arrival, departure)

• How many people stay on Dagobah in movie 3?

SELECT count(*) FROM timetable, characters WHERE movie=3 and pname ='Dagobah' and

timetable.cname=characters.name and characters.race='Human'

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Starwar Exercises

char(name, race, homeworld, affiliation)planets(name, type, affiliation)

timetable(cname, pname, movie, arrival, departure)

• Who has been to his/her homeworld in movie 2?

SELECT distinct c.name FROM characters c, timetable t WHERE c.name=t.cname and t.pname=c.homeworld and

movie=2

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Starwar Exercises

char(name, race, homeworld, affiliation)planets(name, type, affiliation)

timetable(cname, pname, movie, arrival, departure)

• Find all characters that have been on all planets of rebels.

SELECT nameFROM characters c WHERE not exists (

SELECT p.name FROM planets p WHERE affiliation='rebels' and p.name NOT IN

(SELECT pname from timetable t where t.cname=c.name and t.pname=p.name))

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Starwar Exercises

char(name, race, homeworld, affiliation)planets(name, type, affiliation)

timetable(cname, pname, movie, arrival, departure)

• Find distinct names of the planets visited by those of race “droid”.

SELECT distinct t.pname FROM char c, timetable tWHERE c.name=t.cname and c.race='droid'

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Starwar Exercises

char(name, race, homeworld, affiliation)planets(name, type, affiliation)

timetable(cname, pname, movie, arrival, departure)

• For each character and for each neutral planet, how much time total did the character spend on the planet?

SELECT c.name, p.name, SUM(t.departure-t.arrival+1) as amountFROM characters c, timetable t, planets p WHERE t.cname=c.name and t.pname=p.name and p.affiliation='neutral' GROUP BY c.name, p.name